Differentiating Logarithmic Functions with Base e
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Key Questions

Answer:
#1/x# Explanation:
#log_e(x)# is commonly denoted as#ln(x)# , the natural log.#=>d/(dx) ln(x) = 1/x# If you would like a proof, we can derive it from the limit definition:
#lim_(delta x>0)(f(x+delta x)f(x))/(delta x)# #= lim_(delta x>0)(ln(x+delta x)ln(x))/(delta x)# #= lim_(delta x>0)(ln((x+delta x)/(x)))/(delta x)# #= lim_(delta x>0)1/(delta x)ln(1+(delta x)/x)# #= lim_(delta x>0)ln((1+(delta x)/x)^(1/(delta x)))# #= lim_(delta x>0)ln((1+(delta x)/x)^(1/(delta x)))# #"Let " tau equiv (delta x)/x# :#= lim_(delta tau>0)ln((1+tau)^(1/(xtau)))# #= lim_(delta tau>0)ln[((1+tau)^(1/(tau)))^(1/x)]# #= ln[(e)^(1/x)]# #= 1/x ln(e)# #= 1/x (1)# #= 1/x# 
By Quotient Rule,
#y'={1/x cdot xlnx cdot 1}/{x^2}={1lnx}/{x^2}# This problem can also be solved by the Product Rule
#y'=f'(x)g(x)+f(x)g(x)# The original function can also be rewritten using negative exponents.
#f(x)=ln(x)/x=ln(x)*x^1# #f'(x)=1/x*x^1+ln(x)*1x^2# #f'(x)=1/x*1/x+ln(x)*1/x^2# #f'(x)=1/x^2ln(x)/x^2# #f'(x)=(1ln(x))/x^2# 
The answer would be
#f'(x) = 1/g(x)*g'(x)# or it can be written as#f'(x)=(g'(x))/g(x)# .To solve this derivative you will need to follow the chain rule which states:
#F(x)=f(g(x))# then#F'(x)=f'(g(x))*g'(x)# Or without the equation, it the derivative of the outside(without changing the inside), times the derivative of the outside.
The derivative of
#h(x) = ln(x)# is#h'(x) = 1/x# .
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